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3 years ago

Can someone help me solve this problem (graphing/geometry)

Mathematics

1 answer:

PIT_PIT [208]3 years ago

5 0

Answer:

Step-by-step explanation:

i quessed

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What is the solution set of the quadratic inequality 4(x + 2)2 <0?

Lesechka [4]

Step-by-step explanation:

4(X+2)2

taking in left hand side,

(X+2)=4-2

(X+2)=2

X=2-2

X=0

3 0

2 years ago

Dentify on which quadratic function is positive.

Olin [163]

Answer:

$\textsf{$y = 2x^2 - 17x + 30$: \quad $\left(-\infty, \dfrac{5}{2}\right) \cup (6, \infty)$}$

$\textsf{$y = - x^2 - 6x - 8$: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}$

Step-by-step explanation:

A function is positive when it is above the x-axis, and negative when it is below the x-axis.

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Given quadratic equation:

$y = 2x^2 - 17x + 30$

Factor the equation:

$\implies y = 2x^2 - 17x + 30$

$\implies y = 2x^2 - 5x-12x + 30$

$\implies y=x(2x-5)-6(2x-5)$

$\implies y=(x-6)(2x-5)$

The x-intercepts of the parabola are when y = 0.

To find the x-intercepts, set each factor equal to zero and solve for x:

$\implies x-6=0 \implies x=6$

$\implies 2x-5=0 \implies x=\dfrac{5}{2}$

Therefore, the x-intercepts are x = ⁵/₂ and x = 6.

The leading coefficient of the given function is positive, so the parabola opens upwards.

The function is positive when it is above the x-axis.

Therefore, the function is positive for the values of x less than the smallest x-intercept and more than the largest x-intercept:

$\textsf{Solution: \quad $x < \dfrac{5}{2}$ \;and \;$x > 6$}$

$\textsf{Interval notation: \quad $\left(-\infty, \dfrac{5}{2}\right) \cup (6, \infty)$}$

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Given quadratic equation:

$y = - x^2 - 6x - 8$

Factor the equation:

$\implies y = - x^2 - 6x - 8$

$\implies y = -(x^2 +6x +8)$

$\implies y = -(x^2 +4x +2x+8)$

$\implies y = -((x(x+4)+2(x+4))$

$\implies y = -(x+4)(x+2)$

The x-intercepts of the parabola are when y = 0.

To find the x-intercepts, set each factor equal to zero and solve for x:

$\implies x+4=0 \implies x=-4$

$\implies x+2=0 \implies x=-2$

Therefore, the x-intercepts are x = -4 and x = -2.

The leading coefficient of the given function is negative, so the parabola opens downwards.

The function is negative when it is below the x-axis.

Therefore, the function is negative for the values of x less than the smallest x-intercept and more than the largest x-intercept:

$\textsf{Solution: \quad $x < -4$ \;and \;$x > -2$}$

$\textsf{Interval notation: \quad $\left(-\infty, -4\right) \cup (-2, \infty)$}$

4 0

1 year ago

Read 2 more answers

Given the points A(0, 0), B(e, f), C(0, e) and D(f, 0), determine if line segments AB and CD are parallel, perpendicular or neit

Virty [35]

Answer:

Perpendicular

Step-by-step explanation:

Calculate the slopes m of the segments using the slope formula.

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Parallel lines have equal slopes

The product of the slopes of perpendicular lines equals - 1

(x₁, y₁ ) = A(0, 0) and (x₂, y₂ ) = B(e, f)

$m_{AB}$ = $\frac{f-0}{e-0}$ = $\frac{f}{e}$

Repeat with (x₁, y₁ ) = C(0, e) and (x₂, y₂ ) = D(f, 0)

$m_{CD}$ = $\frac{0-e}{f-0}$ = - $\frac{e}{f}$

Thus AB and CD are perpendicular since $\frac{f}{e}$ × - $\frac{e}{f}$ = - 1

5 0

3 years ago

PLEASE HELPPPPPPPPPPPPP

mrs_skeptik [129]

Step-by-step explanation:

The following configurations are evaluated, as given or stated within the interrogate:

P(Q) = 0.6

P(R) = 0.9

When the variable constant of Q and R transition into independent events within the function notation, the product of the individual values, as equated to those particular independent variables, is required.

For example:

If P(Q) = 0.6 and P(R) = 0.9, and Q and R fuse, then find the product of 0.9 and 0.6 is obligated:

0.9 * 0.6 = 0.54

Thus, given the independent events, P(Q and R) is equivalent to 0.54.

3 0

2 years ago

Read 2 more answers

Center: (10,-10)

vova2212 [387]

This is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

What is the ellipse?

The equation for an ellipse is typically written as x² a² + y² b² = 1. x² a² + y² b² = 1. An ellipse with its origin at the center is defined by this equation. The ellipse is stretched further in both the horizontal and vertical directions if a > b, a > b, and if b > a, b > a, respectively.

The standard form of the equation of an ellipse with center (h, k)and major axis parallel to the x-axis is:

$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$

where,

a > b

the length of the major axis is 2a

the coordinates of the vertices are (h±a,k)

the length of the minor axis is 2b

the coordinates of the co-vertices are (h,k±b)

the coordinates of the foci are (h±c,k),

where c² = a² − b².

so,

$\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

Hence, this is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$ .

To learn more about ellipse, visit

brainly.com/question/16904744

#SPJ1

7 0

1 year ago